[LeetCode] 844. Backspace String Compare-优快云博客

本文探讨了在存在退格符的情况下，两个字符串是否相等的问题。通过示例解析了算法原理，并提供了一种O(N)时间复杂度和O(1)空间复杂度的解决方案。

Problem

Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.

Example 1:

Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".

Example 2:

Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T become "".

Example 3:

Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".

Example 4:

Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".

Note:

1 <= S.length <= 200
1 <= T.length <= 200
S and T only contain lowercase letters and '#' characters.

Follow up:

Can you solve it in O(N) time and O(1) space?

Solution

class Solution {
    public boolean backspaceCompare(String S, String T) {
        return helper(S).equals(helper(T));
    }
    private String helper(String str) {
        int count = 0;
        String res = "";
        for (int i = str.length()-1; i >= 0; i--) {
            char ch = str.charAt(i);
            if (ch == '#') count++;
            else {
                if (count > 0) count--; //能不加就不加
                else res += ch; //非加不可的字符 那就加吧
            }
        }
        return res;
    }
}